# Proving a multivariate inequality over $0<x<1$ and $n>2$

EDIT: I meant to have the coefficients reversed, showing: $$\frac{n}{n-1}(1-(1-x)^n)^n + (1-x)^{n-1} \leq 1$$ This version should be true.. but still trying to prove it...

ORIGINAL: Is it possible to show: $$(1-(1-x)^n)^n + \frac{n}{n-1}(1-x)^{n-1} \leq 1$$ for $0<x<1$ and $n\geq 2$ (and $n$ is an integer)? This increases the difficulty over the other questions I just asked.

the second term seems to decrease with $n$, so I can solve for the minimum of $n$. But the first term doesn't completely increase with $n$ -- the lines cross when I plot the first term with $n=2$ and then $n=3$. So I'm not sure where to go from there.

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If $0 < x < 1$ then you can simply replace $y = 1 - x$ and say $0 < y < 1$, making the inequality a bit easier. You could then maybe also replace $z = y^n$ such that again $0 < z < 1$, but I'm not sure that helps. – TMM Sep 14 '11 at 23:11
The $n/(n-1)$ reminds me of Hölder's inequality. – Srivatsan Sep 14 '11 at 23:20
Is $n \geq 2$ or $n > 2$? The title and the (original) question text say different things. – Srivatsan Sep 15 '11 at 13:07
@Srivatsan: Honestly, I would be happy with either! – Angada Sep 15 '11 at 16:34

It fails already for $n=2$. Make the substitution suggested by Thijs, and the left-hand side becomes $(1-y^2)^2+2y = 1+y^4 +2y(1-y) > 1$ for $0<y<1$.

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Thank you!! Turns out, I accidentally wrote the wrong equation, but the comments helped me figure it out anyway. – Angada Sep 15 '11 at 1:23

This might be false for every positive integer $n \ge 2$.

I believe we can show this for $\displaystyle x \gt 1 - \frac{1}{n-1}$ by using Bernoulli's inequality on the first term on the left side.

Using Bernoulli's.

$$(1 - (1-x)^n)^n \ge 1 - n(1-x)^n$$

And so

$$(1 - (1-x)^n)^n + \frac{n(1-x)^{n-1}}{n-1} \ge 1 - n(1-x)^n + \frac{n(1-x)^{n-1}}{n-1}$$

$$= 1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right)$$

if $$1 - x \lt \frac{1}{n-1}$$

then

$$1 - n(1-x)^{n-1}\left((1-x) - \frac{1}{n-1}\right) \gt 1$$

Are you hoping for this to be true, or do you know this to be true (like assigned textbook problem).

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Thank you!! Turns out, I accidentally wrote the wrong equation, so this one was false, but the comments helped me figure it out how to prove the one I meant to write, which I know to be true. – Angada Sep 15 '11 at 1:24